EGMn: The Sequential Endogenous Grid Method
IAAE 2024 – Thessaloniki, Greece
June 26, 2024
Motivation
- Structural Economics for modeling decision-making under uncertainty
- household: consumption, savings, labor, portfolio, retirement
- firms: production, investment, hiring, entry/exit
- governments: fiscal and monetary policy, taxation, redistribution
- interdisciplinary: climate change, public health, education, etc.
- Structural modeling is hard
- modern economics requires solving complex problems
- with many state variables, many decisions, and non-convexities
- computationally challenging and time-consuming
Outline
- Dynamic Programming
- The Endogenous Grid Method
- The Sequential Endogenous Grid Method
- Functional Approximation
- Conventional techniques are insufficient for complex problems
- Neural Nets as function approximators
- Gaussian Process Regression
- Conclusion
- Sequential problems are easier to solve
- GPR is a powerful tool for interpolation on unstructured grids
Dynamic Programming
A simple consumption-savings problem
Agent maximizes present discounted value (PDV) of lifetime utility
\[\begin{equation} \max_{c_t} \sum_{t=0}^{\infty} \beta^t \mathrm{u}(c_t) \end{equation}\]
Recursive Bellman equation
\[\begin{equation} \begin{split} v_t(m_t) & = \max_{c_t} \mathrm{u}(c_t) + \beta\mathbb{E}_t \left[ v_{t+1}(m_{t+1}) \right] \\ \text{s.t.} & \quad 0 < c_t \leq m_t \\ a_t & = m_t - c_t \\ m_{t+1} & = \mathsf{R}a_t + \theta_{t+1} \end{split} \end{equation}\]
A simple consumption-savings problem
Recursive Bellman equation
\[\begin{equation} \begin{split} v_t(m_t) & = \max_{c_t} \mathrm{u}(c_t) + \beta\mathbb{E}_t \left[ v_{t+1}(m_{t+1}) \right] \\ \text{s.t.} & \quad 0 < c_t \leq m_t \\ a_t & = m_t - c_t \\ m_{t+1} & = \mathsf{R}a_t + \theta_{t+1} \end{split} \end{equation}\]
How do we solve this problem?
- Value Function Iteration (VFI)
- Discretize state space (interpolation)
- Grid search optimization (brute force, iterative)
The Endogenous Grid Method
by (Carroll2006-ag?)
\[\begin{equation} v_t(m_t) = \max_{c_t} \mathrm{u}(c_t) + \beta\mathbb{E}_t \left[ v_{t+1}(\mathsf{R}(m_t - c_t) + \theta_{t+1}) \right] \end{equation}\]
\[\begin{equation} u'(c_t) = \beta\mathsf{R}\mathbb{E}_t \left[ v_{t+1}'(\mathsf{R}(m_t - c_t) + \theta_{t+1}) \right] \end{equation}\]
\[\begin{equation} \mathfrak{c}_t([\mathrm{a}]) = \mathrm{u}'^{-1} \left( \beta\mathsf{R}\mathbb{E}_t \left[ v_{t+1}'(\mathsf{R}[\mathrm{a}]+ \theta_{t+1}) \right] \right) \end{equation}\]
\[\begin{equation} \mathfrak{m}_t([\mathrm{a}]) = \mathfrak{c}_t([\mathrm{a}]) + [\mathrm{a}] \end{equation}\]
Contribution: \((\mathfrak{m}_t, \mathfrak{c}_t) \quad \Rightarrow \quad \hat{c}_t(\mathfrak{m}_t) = \mathfrak{c}_t\)
- Simple: Inverted Euler equation
- Fast: No root-finding or grid search optimization required
- Efficient: Finds exact solution at each gridpoint
Limitations of EGM
- One-dimensional problems/sub-problems (nested)
- (GEGM) (Barillas2007?)
- (NEGM) (Druedahl2021?)
- Can result in non-rectangular grids
- (Curvilinear) (White2015?)
- (Triangular) (Ludwig2018?)
- Non-convexities (discrete choices) can be problematic
- (DCEGM) (Iskhakov2017-af?)
- (G2EGM) (Druedahl2017-ac?)
EGMn: The Sequential
Endogenous Grid Method
- Insight: Problems in which agent makes several simultaneous choices can be decomposed into sequence of problems
- Challenge: Rectilinear exogenous grid results in unstructured endogenous grid
- Solution: Use machine learning to interpolate on unstructured grids
Contribution:
- Simple, Fast, Efficient: Inherits properties of EGM
- Multi-dimensional: Can be used for problems with multiple state variables and decisions
- Cutting-edge: Functional approximation and uncertainty quantification approach using Gaussian Process Regression
A consumption - leisure problem
Agent maximizes PDV of lifetime utility
\[\begin{equation} \mathrm{V}_0(B_0, \theta_0) = \max_{C_{t}, Z_{t}} \mathbb{E}_{t} \left[ \sum_{t = 0}^{T} \beta^{t} \mathrm{u}(C_{t}, Z_{t}) \right] \end{equation}\]
Recursive Bellman equation in normalized form:
\[\begin{equation} \begin{split} \mathrm{v}_{t}(b_{t}, \theta_{t}) & = \max_{\{c_{t}, z_{t}\}} \mathrm{u}(c_{t}, z_{t}) + \beta\mathbb{E}_{t} \left[ \Gamma_{t+1}^{1-\rho} \mathrm{v}_{t+1} (b_{t+1}, \theta_{t+1}) \right] \\ \ell_{t} & = 1 - z_{t} \\ m_{t} & = b_{t} + \theta_{t} \mathsf{w}\ell_{t} \\ a_{t} & = m_{t} - c_{t} \\ b_{t+1} & = a_{t} \mathsf{R}/ \Gamma_{t+1} \end{split} \end{equation}\]
A consumption - leisure problem
Recursive Bellman equation in normalized form:
\[\begin{equation} \begin{split} \mathrm{v}_{t}(b_{t}, \theta_{t}) & = \max_{\{c_{t}, z_{t}\}} \mathrm{u}(c_{t}, z_{t}) + \beta\mathbb{E}_{t} \left[ \Gamma_{t+1}^{1-\rho} \mathrm{v}_{t+1} (b_{t+1}, \theta_{t+1}) \right] \\ \ell_{t} & = 1 - z_{t} \\ m_{t} & = b_{t} + \theta_{t} \mathsf{w}\ell_{t} \\ a_{t} & = m_{t} - c_{t} \\ b_{t+1} & = a_{t} \mathsf{R}/ \Gamma_{t+1} \end{split} \end{equation}\]
where
\[\begin{equation} \mathrm{u}(C, Z) = u(C) + h(Z) = \frac{C^{1-\rho}}{1-\rho} + \nu^{1-\rho} \frac{Z^{1-\zeta}}{1-\zeta} \end{equation}\]
Breaking up the problem into sequences
Starting from the beginning of the period, we can define the labor-leisure problem as
\[\begin{equation} \begin{split} \mathrm{v}_{t}(b_{t}, \theta_{t}) & = \max_{ z_{t}} h(z_{t}) + \tilde{\mathfrak{v}}_{t} (m_{t}) \\ & \text{s.t.} \\ 0 & \leq z_{t} \leq 1 \\ \ell_{t} & = 1 - z_{t} \\ m_{t} & = b_{t} + \theta_{t} \mathsf{w}\ell_{t}. \end{split} \end{equation}\]
The pure consumption-saving problem is then
\[\begin{equation} \begin{split} \tilde{\mathfrak{v}}_{t}(m_{t}) & = \max_{c_{t}} u(c_{t}) + \beta\mathfrak{v}_{t}(a_{t}) \\ & \text{s.t.} \\ 0 & \leq c_{t} \leq m_{t} \\ a_{t} & = m_{t} - c_{t}. \end{split} \end{equation}\]
Breaking up the problem into sequences
Starting from the beginning of the period, we can define the labor-leisure problem as
\[\begin{equation} \begin{split} \mathrm{v}_{t}(b_{t}, \theta_{t}) & = \max_{ z_{t}} h(z_{t}) + \tilde{\mathfrak{v}}_{t} (m_{t}) \\ & \text{s.t.} \\ 0 & \leq z_{t} \leq 1 \\ \ell_{t} & = 1 - z_{t} \\ m_{t} & = b_{t} + \theta_{t} \mathsf{w}\ell_{t}. \end{split} \end{equation}\]
The pure consumption-saving problem is then
\[\begin{equation} \begin{split} \tilde{\mathfrak{v}}_{t}(m_{t}) & = \max_{c_{t}} u(c_{t}) + \beta\mathfrak{v}_{t}(a_{t}) \\ & \text{s.t.} \\ 0 & \leq c_{t} \leq m_{t} \\ a_{t} & = m_{t} - c_{t}. \end{split} \end{equation}\]
Finally, the post-decision value function is
\[\begin{equation} \begin{split} \mathfrak{v}_{t}(a_{t}) & = \mathbb{E}_{t} \left[ \Gamma_{t+1}^{1-\rho} \mathrm{v}_{t+1}(b_{t+1}, \theta_{t+1}) \right] \\ & \text{s.t.} \\ b_{t+1} & = a_{t} \mathsf{R}/ \Gamma_{t+1}. \end{split} \end{equation}\]
Solving Labor-Leisure (EGM, Again)
We can condense the labor-leisure problem into a single equation:
\[\begin{equation} \mathrm{v}_{t}(b_{t}, \theta_{t}) = \max_{ z_{t}} h(z_{t}) + \tilde{\mathfrak{v}}_{t}(b_{t} + \theta_{t} \mathsf{w}(1-z_{t})) \end{equation}\]
Interior solution must satisfy the first-order condition:
\[\begin{equation} h'(z_{t}) = \tilde{\mathfrak{v}}_{t}'(m_{t}) \theta_{t} \mathsf{w} \end{equation}\]
EGM consists of inverting the first-order condition to find leisure function:
\[\begin{equation} \mathfrak{z}_{t}([\mathrm{m}], [\mathrm{\theta}]) = h'^{-1}\left( \tilde{\mathfrak{v}}_{t}'([\mathrm{m}]) \mathsf{w}[\mathrm{\theta}]\right) \end{equation}\]
Solving Labor-Leisure (EGM, Again)
Using market resources condition we obtain endogenous grid:
\[\mathfrak{b}_{t}([\mathrm{m}], [\mathrm{\theta}]) = [\mathrm{m}]- [\mathrm{\theta}]\mathsf{w}(1-\mathfrak{z}_{t}([\mathrm{m}], [\mathrm{\theta}]))\]
So now we have the triple \((\mathfrak{z}_t, \mathfrak{b}_t, \theta)\), where \(\mathfrak{z}_t\) is the unconstrained approx. of optimal leisure for each \((\mathfrak{b}_t, \theta)\) corresponding to each \(([\mathrm{m}], [\mathrm{\theta}])\). Generally, we can construct an interpolator as follows:
\[z_t(\mathfrak{b}_t, \theta) = \mathfrak{z}_t\]
Actual leisure function is bounded between 0 and 1:
\[\begin{equation} \hat{z}_{t}(b, \theta) = \max \left[ \min \left[ z_{t}(b, \theta), 1 \right], 0 \right] \end{equation}\]
Pretty Simple, Right?
What is the problem?
|
Exogenous Rectangular Grid |
Endogenous Curvilinear Grid |
- One solution: Curvilinear Interpolation by (White2015?)
Warped Grid Interpolation
- Our solution: Warped Grid Interpolation (simpler, faster, more details on paper)
A more complex problem
Consumption - Pension Deposit Problem as in (Druedahl2017-ac?)
\[\begin{equation} \begin{split} \mathrm{v}_{t}(m_{t}, n_{t}) & = \max_{c_{t}, d_{t}} u(c_{t}) + \beta\mathbb{E}_{t} \left[ \mathrm{v}_{t+1}(m_{t+1}, n_{t+1}) \right] \\ & \text{s.t.} \quad c_{t} > 0, \quad d_{t} \ge 0 \\ a_{t} & = m_{t} - c_{t} - d_{t} \\ b_{t} & = n_{t} + d_{t} + g(d_{t}) \\ m_{t+1} & = a_{t} \mathsf{R}+ \theta_{t+1} \\ n_{t+1} & = b_{t} \mathbf{R}_{t+1} \end{split} \end{equation}\]
where
\[\begin{equation} \mathrm{u}(c) = \frac{c^{1-\rho}}{1-\rho} \qquad \text{and} \qquad \mathrm{g}(d) = \chi\log(1+d). \end{equation}\]
is a tax-advantaged premium on pension contributions.
G2EGM from
(Druedahl2017-ac?)
- If we try to use EGM:
- 2 first order conditions
- multiple constraints difficult to handle
- segments: combinations of first order conditions and constraints
- \(2^{d}\) segments where \(d\) is number of control variables
- requires local triangulation interpolation
Breaking up the problem makes it easier
Consider the problem of a consumer who chooses how much to put into a pension account:
\[\begin{equation} \begin{split} \mathrm{v}_{t}(m_{t}, n_{t}) & = \max_{d_{t}} \tilde{\mathfrak{v}}_{t}(l_{t}, b_{t}) \\ & \text{s.t.} \quad 0 \le d_{t} \le m_t \\ l_{t} & = m_{t} - d_{t} \\ b_{t} & = n_{t} + d_{t} + g(d_{t}) \end{split} \end{equation}\]
After, the consumer chooses how much to consume out of liquid savings:
\[\begin{equation} \begin{split} \tilde{\mathfrak{v}}_{t}(l_{t}, b_{t}) & = \max_{c_{t}} u(c_{t}) + \beta\mathrm{w}_{t}(a_{t}, b_{t}) \\ & \text{s.t.} \quad 0 < c_{t} \le m_t \\ a_{t} & = l_{t} - c_{t} \end{split} \end{equation}\]
Breaking up the problem makes it easier
Consider the problem of a consumer who chooses how much to put into a pension account:
\[\begin{equation} \begin{split} \mathrm{v}_{t}(m_{t}, n_{t}) & = \max_{d_{t}} \tilde{\mathfrak{v}}_{t}(l_{t}, b_{t}) \\ & \text{s.t.} \quad 0 \le d_{t} \le m_t \\ l_{t} & = m_{t} - d_{t} \\ b_{t} & = n_{t} + d_{t} + g(d_{t}) \end{split} \end{equation}\]
After, the consumer chooses how much to consume out of liquid savings:
\[\begin{equation} \begin{split} \tilde{\mathfrak{v}}_{t}(l_{t}, b_{t}) & = \max_{c_{t}} u(c_{t}) + \beta\mathrm{w}_{t}(a_{t}, b_{t}) \\ & \text{s.t.} \quad 0 < c_{t} \le m_t \\ a_{t} & = l_{t} - c_{t} \end{split} \end{equation}\]
And the post-decision value function is defined as:
\[\begin{equation} \begin{split} \mathrm{w}_t(a_t, b_t) & = \mathbb{E}_{t} \left[ \mathrm{v}_{t+1}(m_{t+1}, n_{t+1}) \right] \\ & \text{s.t.} \\ m_{t+1} & = a_{t} \mathsf{R}+ \theta_{t+1} \\ n_{t+1} & = b_{t} \mathbf{R}_{t+1} \end{split} \end{equation}\]
Breaking up the problem makes it easier
Consider the problem of a consumer who chooses how much to put into a pension account:
\[\begin{equation} \begin{split} \mathrm{v}_{t}(m_{t}, n_{t}) & = \max_{d_{t}} \tilde{\mathfrak{v}}_{t}(l_{t}, b_{t}) \\ & \text{s.t.} \quad 0 \le d_{t} \le m_t \\ l_{t} & = m_{t} - d_{t} \\ b_{t} & = n_{t} + d_{t} + g(d_{t}) \end{split} \end{equation}\]
After, the consumer chooses how much to consume out of liquid savings:
\[\begin{equation} \begin{split} \tilde{\mathfrak{v}}_{t}(l_{t}, b_{t}) & = \max_{c_{t}} u(c_{t}) + \beta\mathrm{w}_{t}(a_{t}, b_{t}) \\ & \text{s.t.} \quad 0 < c_{t} \le m_t \\ a_{t} & = l_{t} - c_{t} \end{split} \end{equation}\]
And the post-decision value function is defined as:
\[\begin{equation} \begin{split} \mathrm{w}_t(a_t, b_t) & = \mathbb{E}_{t} \left[ \mathrm{v}_{t+1}(m_{t+1}, n_{t+1}) \right] \\ & \text{s.t.} \\ m_{t+1} & = a_{t} \mathsf{R}+ \theta_{t+1} \\ n_{t+1} & = b_{t} \mathbf{R}_{t+1} \end{split} \end{equation}\]
Steps:
- Compute \(\mathrm{w}_t(a_t, b_t)\)
- Solve consumption problem (EGM)
- Solve pension problem (EGM, again)
- Done!
Solving the pension problem
The pension problem, more compactly
\[\begin{equation} \mathrm{v}_{t}(m_{t}, n_{t}) = \max_{d_{t}} \tilde{\mathfrak{v}}_{t}(m_{t} - d_{t}, n_{t} + d_{t} + \mathrm{g}(d_{t})) \end{equation}\]
Interior solution must satisfy the first-order condition:
\[\begin{equation} \mathrm{g}'(d_{t}) = \frac{\tilde{\mathfrak{v}}_{t}^{l}(l_{t}, b_{t})}{\tilde{\mathfrak{v}}_{t}^{b}(l_{t}, b_{t})} - 1 \end{equation}\]
Inverting, we can obtain the optimal choice of \(d_{t}\):
\[\begin{equation} \mathfrak{d}_{t}(l_{t}, b_{t}) = \mathrm{g}'^{-1}\left( \frac{\tilde{\mathfrak{v}}_{t}^{l}(l_{t}, b_{t})}{\tilde{\mathfrak{v}}_{t}^{b}(l_{t}, b_{t})} - 1 \right) \end{equation}\]
Solving the pension problem
Inverting, we can obtain the optimal choice of \(d_{t}\):
\[\begin{equation} \mathfrak{d}_{t}(l_{t}, b_{t}) = \mathrm{g}'^{-1}\left( \frac{\tilde{\mathfrak{v}}_{t}^{l}(l_{t}, b_{t})}{\tilde{\mathfrak{v}}_{t}^{b}(l_{t}, b_{t})} - 1 \right) \end{equation}\]
Using resource constraints we obtain endogenous grids:
\[\begin{equation} \mathfrak{n}_{t}(l_{t}, b_{t}) = b_{t} - \mathfrak{d}_{t}(l_{t}, b_{t}) - \mathrm{g}(\mathfrak{d}_{t}(l_{t}, b_{t})) \\ \mathfrak{m}_{t}(l_{t}, b_{t}) = l_{t} + \mathfrak{d}_{t}(l_{t}, b_{t}) \end{equation}\]
Now we have the triple \(\{\mathfrak{m}_t, \mathfrak{n}_t, \mathfrak{d}_t\}\) where \(\mathfrak{d}_t\) is the unconstrained approx. of optimal deposit for each \((\mathfrak{m}_t, \mathfrak{n}_t)\) corresponding to each \((l_t, b_t)\). Generally, we can construct an interpolator as follows:
\[\begin{equation} \hat{d_t}(\mathfrak{m}_t, \mathfrak{n}_t) = \begin{cases} 0 & \text{if } \mathfrak{d}_t < 0 \\ \mathfrak{d}_t & \text{if } 0 \le \mathfrak{d}_t \le \mathfrak{m}_t \\ \mathfrak{m}_t & \text{if } \mathfrak{d}_t > \mathfrak{m}_t \end{cases} \end{equation}\]
Unstructured Grids
Problem: Rectilinear exogenous grid results in unstructured endogenous grid
Exogenous Rectangular Grid
Endogenous Unstructured Grid
How do we interpolate on this grid?
Functional Approximation
Linear Interpolation on a Uniform Grid
Linear Interpolation on a Non-linear Grid
Bilinear Interpolation
Curvilinear (Warped) Grid Interpolation
See: (White2015?)
What about Unstructured Grids?
See: (Ludwig2018?)
Artificial Neural Networks
Artificial Neural Networks
- Based on biological neural pathways (neurons in a brain)
- Learns function \(f(X): R^n \rightarrow R^m\)
- Consists of
- input (features) \(X\)
- hidden layers \(g(\cdots)\)
- output (target) \(y = f(X)\)
- Hidden layers can have many nodes
- Neural nets can have many hidden layers (deep learning)
A single neuron, and a bit of math
Figure 2: Perceptron
\[\begin{equation} y = g(w_0 + \sum_{i=1}^n w_i x_i) = g(w_0 + \mathbf{x}' \mathbf{w}) \end{equation}\]
A single neuron, and a bit of math
\[\begin{equation} y = g(w_0 + \sum_{i=1}^n w_i x_i) = g(w_0 + \mathbf{x}' \mathbf{w}) \end{equation}\]
- \(y\) is the output or target
- \(x_i\) are the inputs or features
- \(w_0\) is the bias
- \(w_i\) are the weights
- \(g(\cdot)\) is the activation function (non-linear) \[\begin{equation} g(z) = \frac{1}{1 + e^{-z}} \end{equation}\]
- usually a sigmoid, but there are many others
Training a Neural Network
Mean Squared Error (MSE)
\[\begin{equation} J(\mathbf{w}) = \frac{1}{2n} \sum_{i=1}^n \left[y_i - f(\mathbf{x}^{(i)}; \mathbf{w})\right]^2 \end{equation}\]
Objective
\[\begin{equation} \mathbf{w}^* = \arg \min_{\mathbf{w}} J(\mathbf{w}) \end{equation}\]
Training a Neural Network
Mean Squared Error (MSE)
\[\begin{equation} J(\mathbf{w}) = \frac{1}{2n} \sum_{i=1}^n \left[y_i - f(\mathbf{x}^{(i)}; \mathbf{w})\right]^2 \end{equation}\]
Objective
\[\begin{equation} \mathbf{w}^* = \arg \min_{\mathbf{w}} J(\mathbf{w}) \end{equation}\]
Gradient Descent
\[\begin{equation} \mathbf{w}^{(t+1)} = \mathbf{w}^{(t)} - \eta \nabla J(\mathbf{w}^{(t)}) \end{equation}\]
Training a Neural Network
Mean Squared Error (MSE)
\[\begin{equation} J(\mathbf{w}) = \frac{1}{2n} \sum_{i=1}^n \left[y_i - f(\mathbf{x}^{(i)}; \mathbf{w})\right]^2 \end{equation}\]
Objective
\[\begin{equation} \mathbf{w}^* = \arg \min_{\mathbf{w}} J(\mathbf{w}) \end{equation}\]
Stochastic Gradient Descent
\[\begin{equation} \widetilde{\nabla} J(\mathbf{w}^{(t)}) = \frac{1}{B} \sum_{i=1}^B \nabla J_i(\mathbf{w}^{(t)}) \end{equation}\]
\[\begin{equation} \mathbf{w}^{(t+1)} = \mathbf{w}^{(t)} - \eta \widetilde{\nabla} J(\mathbf{w}^{(t)}) \end{equation}\]
Gaussian Process Regression
A Gaussian Process is a probability distribution over functions
\[\begin{equation} \begin{gathered} f(x) \sim \mathcal{GP}(m(x), k(x, x')) \\ \text{where} \quad m(x) = \mathbb{E}[f(x)] \\ \text{and} \quad k(x, x') = \mathbb{E}[(f(x) - m(x))(f(x') - m(x'))] \end{gathered} \end{equation}\]
A Gaussian Process Regression is used to find the function that best fits a set of data points
\[\begin{equation} \mathbb{P}(\mathbf{f} | \mathbf{X}) = \mathcal{N}(\mathbf{f} | \mathbf{m}, \mathbf{K}) \end{equation}\]
I use standard covariance function, exploring alternatives is an active area of research
\[\begin{equation} k(\mathbf{x}_i, \mathbf{x}_j) = \sigma^2_f \exp\left(-\frac{1}{2l^2} (\mathbf{x}_i - \mathbf{x}_j)' (\mathbf{x}_i - \mathbf{x}_j)\right). \end{equation}\]
Approximation
Note
Gaussian Processes
- mathematically equivalent to neural networks with an infinite width
- do not require as much data as neural networks
- offers uncertainty quantification of the mean function
- can be used to approximate any function arbitrarily closely
Universal Approximation Theorem
A single hidden-layer ANN can approximate any continuous function arbitrarily closely as the number of neurons in the hidden layer increases.
An example
Consider the true function \(f(x) = x \cos(1.5x)\) sampled at random points
True Function
An example
A random sample of the GP posterior distribution of functions
Posterior Sample
An example
Gaussian Process Regression finds the function that best fits the data
- Gaussian Process Regression gives us
- Mean function of the posterior distribution
- Uncertainty quantification of the mean function
- Can be useful to predict where we might need more points and update the grid
Back to the model
Second Stage Pension Endogenous Grid
Some Results
Consumption Function
Deposit Function
Conclusion
Conditions for using Sequential EGM
- Model must be
- concave
- differentiable
- continuous
- separable
Need an additional function to exploit invertibility
Examples in this paper:
- Separable utility function
- \(\mathrm{u}(c, z) = \mathrm{u}(c) + h(z)\)
- Continuous and differentiable transition
- \(b_{t} = n_{t} + d_{t} + g(d_{t})\)
Resources
- An Introduction to Statistical Learning
statlearning.com - Neural Networks and Deep Learning
neuralnetworksanddeeplearning.com - Deep Learning
deeplearningbook.org - Probabilistic machine learning
probml.github.io/pml-book - A Neural Network Playground
playground.tensorflow.org - The Gaussian Process Web Site
gaussianprocess.org - A Visual Exploration of Gaussian Processes
distill.pub/2019 - Interactive Gaussian Process Visualization
http://www.infinitecuriosity.org/vizgp
Thank you!
engine: github.com/econ-ark/HARK
References
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